Improper Integral Question/Check

In summary, the person is discussing their homework in improper integrals for their real analysis class. They have solved two problems, one involving log(x) and the other involving log(x)/x, but are unsure if they have made a mistake as their calculator gives a different answer. They are also discussing the differences between log_e and log_10 in mathematics and science.
  • #1
moo5003
207
0
Hello, I just finished up two problems for my homework and I have a sneaking notion that I have made a mistake somewhere because when I checked the answer numerically by calculator and I get a differing number.

I'm doing improper integrals for my real analysis class and the problem is stated as:

Calculate a) Integral from 0 to 1 of log(x)

Work: Derivative of (xlog(x)-x) = log(x) thus: let 0<d<1

Lim d-> 0 (from the right) Integral from d to 1 of log(x) = Lim d -> 0 xlog(x)-x evaluated from d to 1 = lim d -> 0 of d-1-dlog(d) = 0 - 1 - 0*-Infinity = -1

Integral from 0 to 1 of Log(x) = -1

Though, when i do this on my calculator i get something around -.43.

Similarly I get:

Integral from 2 to Infinity of Log(x)/x = -log^2(2)/2
Integral from 0 to Infinity of 1/(x^2+1) = Undefined (since Lim x->Infinity of Tan^-1(x) does not exist).

For part b/c. Thanks for anyhelp you can provide in finding my mistake at least for part a so I can recheck part b/c.:smile:
 
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  • #2
a) This looks right. You should get the same number as when you look at

[tex]-\int_{-\infty}^0 e^x dx[/tex]

Do you see why?

b)What steps did you go through here?

c) I don't know whether this is undefined, but I do know that

[tex] \lim_{x\rightarrow\infty}\arctan (x) = \frac{\pi}{2} [/tex]
 
  • #3
moo5003 said:
Though, when i do this on my calculator i get something around -.43.

Log on your calculator is log_10, not log_e, and

[tex] \int_0^1 log_{10} x dx = -0.43.[/itex]

In math, the default interpretation of log is log_e; in science, log often means log_10.
 

1. What is an improper integral?

An improper integral is an integral where one or both of the limits of integration are infinite, or the integrand is unbounded. It essentially represents the area under a curve that extends to infinity.

2. How do you determine if an improper integral converges or diverges?

If the limit of the integral exists, then the improper integral converges. If the limit does not exist, then the improper integral diverges. This can be determined by evaluating the integral as the upper limit approaches infinity or the lower limit approaches negative infinity.

3. What are some common techniques for evaluating improper integrals?

Some common techniques include using integration by parts, using trigonometric substitutions, and breaking the integral into smaller pieces that can be evaluated separately. In some cases, it is also helpful to rewrite the integrand in terms of partial fractions.

4. Can an improper integral have a finite limit but still be divergent?

Yes, it is possible for an improper integral to have a finite limit but still be divergent. This occurs when the integral has an infinite oscillation or a discontinuity at one or both of the limits of integration.

5. How can improper integrals be applied in real-life situations?

Improper integrals have various applications in physics, engineering, and economics. For example, they can be used to calculate the total distance traveled by an object with changing velocity, the total amount of fluid flowing through a pipe with varying flow rate, or the total profit earned by a company with fluctuating revenue.

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